Question

The increase in pressure required to decrease the volume of 200 L of water by 0.004 percent is (Bulk modulus of water is \( 2.1 \times 10^9 \) N/m\(^2\)):

• A. \( 8.4 \times 10^4 \) N/m\(^2\)
• B. \( 8.4 \times 10^3 \) N/m\(^2\)
• C. \( 8.4 \times 10^5 \) N/m\(^2\)
• D. \( 8.4 \times 10^6 \) N/m\(^2\)

Hint:

The bulk modulus describes how much pressure is needed to compress a substance. The greater the bulk modulus, the more pressure is required for a given volume change.

Answer 1

The Correct Option is A

Step 1: Convert the percentage volume change to a decimal fraction: \( 0.004\% = \dfrac{0.004}{100} = 4\times10^{-5} \).

Step 2: Use the bulk modulus relation \( K = \dfrac{\Delta P}{\Delta V/V} \), rearranged as \( \Delta P = K \times (\Delta V/V) \).

Step 3: Substitute \( K = 2.1\times10^{9}\ \text{N/m}^2 \) and the fraction from Step 1: \( \Delta P = 2.1\times10^{9}\times4\times10^{-5} \).

Step 4: Multiply out: \( 2.1\times4=8.4 \), and the powers of ten give \( 10^{9-5}=10^{4} \).\[ \boxed{\Delta P = 8.4\times10^{4}\ \text{N/m}^2} \]

Answer 2

Another way to see this is to express the volume change in parts per million instead of a percentage. Since \( 1\% = 10{,}000 \) ppm, \( 0.004\% \) converts to \( 0.004 \times 10{,}000 = 40 \) ppm, meaning the volume shrinks by 40 millionths of its original value: \( \Delta V/V = 40\times10^{-6} \).

Applying the bulk modulus relation:

\[ \Delta P = K \times \frac{\Delta V}{V} = (2.1\times10^{9})\times(40\times10^{-6}) = 8.4\times10^{4}\ \text{N/m}^2 \]

1. Option A, \( 8.4\times10^{4} \): agrees with the ppm-based calculation and is correct.
2. Option B, \( 8.4\times10^{3} \): would require the fractional change to be only 4 ppm, ten times smaller than what 0.004% actually represents.
3. Option C, \( 8.4\times10^{5} \): would require 400 ppm, ten times larger than 0.004%.
4. Option D, \( 8.4\times10^{6} \): would require 4000 ppm, a hundred times larger than the given percentage.

The correct answer is \( 8.4\times10^{4}\ \text{N/m}^2 \).